The problem in the image asks to "Compute the value of $\sum_{n=1}^{100} \left( 2^n - 3^n \right)$, given $\sum_{n=1}^{k} x^n = \frac{x(x^k - 1)}{x - 1}$ for any integer $k \geq 1$ and $x \neq 1$."

Solution:

We are required to compute the sum: $\sum_{n=1}^{100} \left( 2^n - 3^n \right)$ This can be split into two separate sums: $\sum_{n=1}^{100} 2^n - \sum_{n=1}^{100} 3^n$ Let's compute each sum separately using the formula provided: $\sum_{n=1}^{k} x^n = \frac{x(x^k - 1)}{x - 1}$ For $x = 2$ and $k = 100$: $\sum_{n=1}^{100} 2^n = \frac{2(2^{100} - 1)}{2 - 1} = 2(2^{100} - 1) = 2^{101} - 2$ For $x = 3$ and $k = 100$: $\sum_{n=1}^{100} 3^n = \frac{3(3^{100} - 1)}{3 - 1} = \frac{3(3^{100} - 1)}{2} = \frac{3 \times 3^{100} - 3}{2} = \frac{3^{101} - 3}{2}$ Thus, the value of the original sum is: $\sum_{n=1}^{100} \left( 2^n - 3^n \right) = \left( 2^{101} - 2 \right) - \frac{3^{101} - 3}{2}$ Let's simplify this expression: $= 2^{101} - 2 - \frac{3^{101}}{2} + \frac{3}{2}$ Finally, the simplified expression is: $\boxed{2^{101} - \frac{3^{101}}{2} - 2 + \frac{3}{2}}$

Do you want any further details on this problem or have any questions?

Here are five related questions for practice:

1. How would the solution change if the upper limit was 50 instead of 100?
2. What would be the sum if $x = 4$ instead of 2 in the given summation formula?
3. Can you derive the general form for $\sum_{n=1}^{k} x^n$ using the geometric series formula?
4. What happens to the sum if $x = 1$?
5. How does the sum behave as $k$ approaches infinity?

Tip: When dealing with sums involving powers, always check if a geometric series formula or a known summation identity can simplify the calculation.